Worked examples — R-squared and adjusted R-squared

Before we start, let us re-anchor the four quantities every example below leans on, in plain words, so nothing uses a symbol you have not seen.

Recall The four quantities, in words (click to reveal)

What is $\bar{y}$? ::: The plain average of the true values — the number you'd guess if you knew nothing else. What is $S{S}_{tot}$? ::: Add up, for every point, how far the true value sits from that average, squared. It measures how spread-out the data is to begin with. What is $S{S}_{res}$? ::: Add up how far each true value sits from the model's guess, squared. It measures the error that is left over after the model tried. What is $R^2$? ::: $1-\frac{S{S}_{res}}{S{S}_{tot}}$ — the fraction of the original spread the model managed to erase.

We square each gap so that a miss of $+10$ and a miss of $-10$ both count as badness, never cancelling. The units of every $SS$ are "(units of $y$) squared", so when we divide one $SS$ by another the units cancel and $R^2$ is a pure number — that is why it reads like a percentage.

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Where the adjusted "rent" comes from

Before Ex 5, one gap from the parent note deserves filling: why is the penalty factor exactly $\frac{n-1}{n-p-1}$?

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The scenario matrix

Every question about $R^2$ lives in one of these cells. Each example below is tagged with the cell it fills.

Cell	What makes it special	Expected behaviour	Example
A. Perfect fit	$S{S}_{res}=0$	$R^2=1$	Ex 1
B. Typical good fit	small $S{S}_{res}$	$0<R^2<1$	Ex 2
C. Model = mean	$S{S}_{res}=S{S}_{tot}$	$R^2=0$	Ex 3
D. Worse than mean	$S{S}_{res}>S{S}_{tot}$	$R^2<0$	Ex 4
E. Adjusted penalty helps	useless feature added	$R^2\uparrow$ but $R_{adj}^2\downarrow$	Ex 5
F. Degenerate d.o.f.	$p\to n-1$	$R_{adj}^2$ collapses	Ex 6
G. Real-world word problem	pick a model	reasoning + numbers	Ex 7
H. Exam twist	$R^2$ from correlation $r$	$R^2=r^2$ (simple reg.)	Ex 8
Z. Degenerate target	$S{S}_{tot}=0$	$R^2$ undefined	callout above

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Ex 1 — Cell A: the perfect fit

Forecast: guess the answer before reading on. (It is the largest value $R^2$ can take.)

1. Mean. $\bar{y}=\frac{3+6+9}{3}=6$. Why this step? The mean is the baseline every $R^2$ compares against — we always need it first.
2. Total spread. $S{S}_{tot}=(3-6{)}^2+(6-6{)}^2+(9-6{)}^2=9+0+9=18$. Why this step? This is the amount of variation the model is being asked to explain.
3. Leftover error. $S{S}_{res}=(3-3{)}^2+(6-6{)}^2+(9-9{)}^2=0$. Why this step? Predictions equal the truth, so every gap is zero.
4. Assemble. $R^2=1-\frac{0}{18}=1$.

Verify: $R^2=1$ is the ceiling — it can only happen when $S{S}_{res}=0$, exactly our case. ✓

Figure 1 — Ex 1: mint dots sit exactly on the lavender "predicted = actual" diagonal, so there are no vertical error gaps to add up. Observe that zero vertical distance is what forces $S{S}_{res}=0$ and $R^2=1$.

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Ex 2 — Cell B: the typical good fit

Forecast: the errors are small but non-zero, so guess a number close to 1 but below it.

1. Mean. $\bar{y}=\frac{10+20+30+40}{4}=25$. Why this step? The mean is our do-nothing baseline; $R^2$ measures how far past it the model gets.
2. Total spread. $S{S}_{tot}=(10-25{)}^2+(20-25{)}^2+(30-25{)}^2+(40-25{)}^2=225+25+25+225=500$. Why this step? Records the raw variation before the model helps.
3. Leftover error. $S{S}_{res}=(10-12{)}^2+(20-18{)}^2+(30-33{)}^2+(40-37{)}^2=4+4+9+9=26$. Why this step? The squared misses that survive the model.
4. Assemble. $R^2=1-\frac{26}{500}=1-0.052=0.948$.

Verify: $0<0.948<1$, and since our errors are tiny compared to the spread, a near-1 value is exactly what we forecast. ✓

Figure 2 — Ex 2: coral error bars (truth → prediction) are short next to the lavender bars (truth → mean). Because the coral total (26) is tiny against the lavender total (500), the model erased almost all the spread → $R^2\approx 0.95$.

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Ex 3 — Cell C: model is no better than the mean

Forecast: if the model literally is the mean, how much extra variation did it explain? Guess the floor.

1. Mean. $\bar{y}=\frac{4+8+12}{3}=8$. Why this step? We need the baseline first — and here the model happens to equal it, which is the whole point.
2. Total spread. $S{S}_{tot}=(4-8{)}^2+(8-8{)}^2+(12-8{)}^2=16+0+16=32$.
3. Leftover error. $S{S}_{res}=(4-8{)}^2+(8-8{)}^2+(12-8{)}^2=16+0+16=32$. Why this step? Because $\hat{y}=\bar{y}$ everywhere, this sum is identical to $S{S}_{tot}$.
4. Assemble. $R^2=1-\frac{32}{32}=0$.

Verify: $S{S}_{res}=S{S}_{tot}\Rightarrow R^2=0$. A model that always guesses the mean is the definition of "explains nothing beyond the baseline". ✓

Figure 3 — Ex 3: the flat lavender line IS the mean, so the model's error bars land exactly on the mean's error bars. Identical bars ⇒ $S{S}_{res}=S{S}_{tot}$ ⇒ $R^2=0$.

This is the boundary between error-based metrics and the baseline: any real model must beat this to be worth keeping.

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Ex 4 — Cell D: worse than the mean → negative R-squared

Forecast: the predictions are wildly off. Can $R^2$ go below its "floor" of 0? Guess yes or no.

1. Mean. $\bar{y}=\frac{4+8+12}{3}=8$. Why this step? We still need the baseline to compare against — even a terrible model is judged relative to the mean.
2. Total spread. $S{S}_{tot}=(4-8{)}^2+(8-8{)}^2+(12-8{)}^2=32$.
3. Leftover error. $S{S}_{res}=(4-30{)}^2+(8-30{)}^2+(12-30{)}^2=676+484+324=1484$. Why this step? The model's guesses are far worse than the mean, so its leftover error explodes.
4. Assemble. $R^2=1-\frac{1484}{32}=1-46.375=-45.375$.

Verify: $S{S}_{res}>S{S}_{tot}$, so $\frac{S{S}_{res}}{S{S}_{tot}}>1$, so $R^2<0$. Negative $R^2$ is not a bug — it is the metric shouting "you'd have done better predicting the average". This is exactly Common Mistake 3 from the parent note. ✓

Figure 4 — Ex 4: the coral prediction line sits far above the mint data cloud, while the lavender mean line hugs it. The coral error gaps dwarf the spread around the mean — that overshoot is exactly what pushes $R^2$ below zero.

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Ex 5 — Cell E: adjusted R-squared catches a useless feature

Forecast: $R^2$ rose. Guess whether the adjusted score rose or fell.

Recall the penalty form (derived in the "rent" section above): $R_{adj}^2=1-(1-R^2)\frac{n-1}{n-p-1}$.

1. Before ($p=2$). $R_{adj}^2=1-(1-0.90)\frac{10-1}{10-2-1}=1-0.10\cdot \frac{9}{7}$. Why this step? We divide the leftover-fraction $0.10$ by the degrees of freedom to charge rent for each parameter.
2. Compute: $\frac{9}{7}=1.2857\dots$, so $R_{adj}^2=1-0.128571=0.871429$.
3. After ($p=3$). $R_{adj}^2=1-(1-0.905)\frac{9}{10-3-1}=1-0.095\cdot \frac{9}{6}$. Why this step? One more feature means the denominator shrank from 7 to 6 — the rent went up.
4. Compute: $\frac{9}{6}=1.5$, so $R_{adj}^2=1-0.1425=0.8575$.

Verify: $R^2$ went $0.90\to 0.905$ (up), but $R_{adj}^2$ went $0.8714\to 0.8575$ (down). The adjusted metric correctly flagged the noise feature — this is why we prefer it during Feature Selection. ✓

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Ex 6 — Cell F: degenerate degrees of freedom

Forecast: almost as many features as data points. Guess: does $R_{adj}^2$ stay near 0.99 or crash?

1. Degrees of freedom left. $n-p-1=50-48-1=1$. Why this step? This tiny number is the alarm — only 1 "free" data point remains after fitting.
2. Apply. $R_{adj}^2=1-(1-0.99)\frac{50-1}{1}=1-0.01\cdot 49$. Why this step? Dividing by 1 means the penalty factor 49 is applied at full force.
3. Compute: $R_{adj}^2=1-0.49=0.51$.

Verify: the impressive $R^2=0.99$ collapses to $0.51$. As $p\to n-1$, the denominator $n-p-1\to 1$ and the penalty balloons — a mathematical signature of Overfitting. If $p=n-1$ exactly, the denominator hits 0 and $R_{adj}^2$ is undefined (division by zero). ✓

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Ex 7 — Cell G: real-world word problem

Forecast: the predictions look close. Guess an $R^2$ above or below 0.9.

1. Mean. $\bar{y}=\frac{220+240+260+250+300+330}{6}=\frac{1600}{6}=266.667$ dollars. Why this step? Baseline: "just predict the typical day's revenue".
2. Total spread. deviations from $266.667$ are $[-46.667,-26.667,-6.667,-16.667,33.333,63.333]$. $S{S}_{tot}=2177.8+711.1+44.4+277.8+1111.1+4011.1=8333.33$ (dollars²). Why this step? Measures how much revenue naturally bounces around.
3. Leftover error. $S{S}_{res}=(220-230{)}^2+(240-235{)}^2+(260-255{)}^2+(250-260{)}^2+(300-295{)}^2+(330-320{)}^2$ $=100+25+25+100+25+100=375$ (dollars²). Why this step? The revenue the model still misses.
4. Assemble. $R^2=1-\frac{375}{8333.33}=1-0.045=0.955$.

Verify: units: dollars²/dollars² cancel → pure number, good. $0.955$ means the model erases about 95.5% of the revenue variation. Above 0.9, as forecast. The owner should still confirm on new days via Cross-Validation before trusting it — training $R^2$ alone never proves generalisation. ✓

Figure 5 — Ex 7: daily revenue (mint) vs model (coral) across six days; the dashed lavender line is the mean baseline. The coral curve tracks the ups and downs closely while the flat baseline misses them — that tracking is the visual meaning of $R^2=0.955$.

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Ex 8 — Cell H: exam twist, R-squared from correlation

Forecast: guess the sign of $R^2$ before reading.

1. Recall the identity. For simple linear regression, $R^2=r^2$ (stated in the parent note). Why this step? This shortcut only holds with one feature — it is the exam's trap to test whether you know that.
2. Square it. $R^2=(-0.8{)}^2=0.64$. Why this step? Squaring destroys the sign — a negative correlation still explains variance.
3. Address the misconception. $r$ tells you the direction of the relationship; $R^2$ tells you the strength of the fit. A strong negative trend fits just as well as a strong positive one.

Verify: $R^2=0.64\ge 0$, contradicting the student. A negative $r$ never makes $R^2$ negative because we square it. (The $R^2=r^2$ shortcut fails for multiple regression — do not carry it over.) ✓

Figure 6 — Ex 8: a downward-sloping cloud has $r=-0.8$ (negative direction) yet a tight fit. Squaring folds the negative sign away, so $R^2=0.64$ — direction and strength are different questions.

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Recall Self-test (click to reveal)

If $S{S}_{res}=S{S}_{tot}$, what is $R^2$? ::: Exactly 0 — the model matched the mean and nothing more. Can ordinary least squares with an intercept give a negative $R^2$ on training data? ::: No — never. With an intercept the fitted line always beats or ties the mean on the training set, so training $R^2\ge 0$ is guaranteed. Negative $R^2$ needs a dropped intercept, a bad manual model, or held-out test data. Adding a noise feature: $R^2$ does what? $R_{adj}^2$ does what? ::: $R^2$ stays same or rises; $R_{adj}^2$ typically falls. What happens to $R^2$ when all $y$ are identical? ::: $S{S}_{tot}=0$, so the formula divides by zero and $R^2$ is undefined. Simple regression with $r=0.5$: $R^2=?$ ::: $0.25$.

See also: Bias-Variance Tradeoff, Model Selection, Linear Regression, 2.2.08 R-squared and adjusted R-squared (Hinglish).